Question

At a certain temperature, K for the reaction below is 7.5 L/mol.

2NO₂(g) → N₂O₄(g)

If 2.0 moles of NO₂ are placed in a 2.0 liter container and permitted to react at this temperature, what will be the concentration of N₂O₄ at equilibrium?

Answer 1

Answer –

Given, K = 7.5 L/mol, moles of NO₂ = 2.0 moles, volume = 2.0 L

We need to calculate the concentration of N₂O₄ at equilibrium

Step 1) Calculate the molarity of NO₂

We know,

Molarity = moles / L

               = 2.0 mole / 2.0 L = 1 M

Step2) We need to put ICE chart

   2NO₂(g) -----> N₂O₄(g)

I 1 M                    0

C -2x                   +x

E 1-2x                  +x

So, K = [N₂O₄(g)] / [NO₂(g)]²

7.5 L/mol = x / (1-2x)²

7.5 *(1-2x)² = x

7.5 (4x²-4x+1 ) = x

30x²-30x+7.5 =x

So, 30x²-30x-x+7.5=0

30x²-31x+7.5 = 0

Now using the quadratic equation

a= 30 , b = -31 , c = 7.5

x = -b +/-√b²-4ac /2a

by plugging the value in this one

x =0.386 and x = 647

we consider the lowest value so,

x = 0.386

we know at equilibrium

x = [N₂O₄(g)] = 0.386 M